Use the difference of cubes to rewrite the expression.
#(a-b)(a^2+ab+b^2) = a^3-b^3#
So,
#(root3a-root3b)(root3a^2+root3(ab)+root3b^2) = a-b#
And, so
#(root3a-root3b)/1 * (root3a^2+root3(ab)+root3b^2)/(root3a^2+root3(ab)+root3b^2) = (a-b)/(root3a^2+root3(ab)+root3b^2)#
In this case, we have:
#root3[x^3+2] - root3[x^3-1] = ((x^3+2)-(x^3-1))/(root3(x^3+2)^2+root3((x^3+2)(x^3-1))+root3(x^3-1)^2)#
# = 3/(root3(x^3+2)^2+root3((x^3+2)(x^3-1))+root3(x^3-1)^2)#
As #x# increases without bound, the denominator also increases without bound, so the limit of the ratio is #0#.